3(2k-1)-8=4(k+1)+5

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Solution for 3(2k-1)-8=4(k+1)+5 equation: 



3(2k-1)-8=4(k+1)+5

We move all terms to the left:

3(2k-1)-8-(4(k+1)+5)=0

We multiply parentheses

6k-(4(k+1)+5)-3-8=0



We calculate terms in parentheses: -(4(k+1)+5), so:

4(k+1)+5

We multiply parentheses

4k+4+5

We add all the numbers together, and all the variables

4k+9

Back to the equation:

-(4k+9)



We add all the numbers together, and all the variables

6k-(4k+9)-11=0

We get rid of parentheses

6k-4k-9-11=0

We add all the numbers together, and all the variables

2k-20=0

We move all terms containing k to the left, all other terms to the right

2k=20

k=20/2

k=10

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